Physics Challenge II: Bouncing out of the atmosphere

[micromass]

Part 1:
Consider $n$ balls $B_1$, $B_2$, ..., $B_n$ having masses $m_1$, ..., $m_n$, such that $m_1\ll m_2\ll ...\ll m_n$. The $n$ balls are stacked above each other. The bottom of $B_1$ is a height $h$ above the ground, and the bottom $B_n$ is a height $\ell$ above the ground. The balls are dropped. In terms of $n$, to what height does the top ball bounce. Assume the balls bounce elastically, assume there is no wind resistance, etc.

Part 2: Assume that $h$ is $1$ meter and assume $\ell$ is negligible. How big should $n$ be in order that the top one reaches a height of $1$ kilometer. How big should $n$be in order that the top one reaches escape velocity? (Escape velocity is approximately $11200~m/s$)

 

[jbriggs444]

Seems trivial enough...

Treat the collision of the stack with the ground as a series of collisions. Ball 1 with the ground, ball 2 with ball 1, etc.

Let v be the velocity of the stack when the first ball strikes the ground.
Let v_n denote the recoil velocity of each ball after colliding with the previous.

Because each collision is elastic and because each ball is much less massive than the previous, the center-of-momentum frame is approximately co-moving with the previous ball. In this frame, the impact velocity of ball i will be v + v_i-1. Its recoil velocity will be identical.

In the ground frame, the recoil velocity v_i = v + 2v_i-1. This is a first order linear recurrence relation marred by the fact that it is inhomogeneous. In the case at hand, we can easily compute v and can easily run an iteration to compute the first ten or twenty values of v_i.

In meters per second,

v = sqrt(2gh) = sqrt(19.6) ~= 4.42 meters/sec.

0 0 << Ball 0 is "the ground"
1 4.42
2 13.26
3 30.94
4 66.3
5 137.02 << Just under enough to reach 1 km
6 278.46 << Enough to reach 1 km.
7 561.34
8 1127.1
9 2258.62
10 4521.66
11 9047.74
12 18099.9 << Escape velocity exceeded by ball 12
13 36204.22
14 72412.86
15 144830.14
16 289664.7
17 579333.82
18 1158672.06
19 2317348.54

But this is a math forum. Numerical solutions are not as classy as analytical solutions.

I am rather rusty at solving recurrence relations and, in particular, in doing the change of variables that is required to make them homogeneous -- it has been about 35 years since we covered that material in school.

v_i+1 = v + 2*v_i
v_i+1 + v = 2 * ( v + v_i )

A change of variables to let v'_i = v_i + v should work.

So v'_i = v'₀ * 2ⁱ

The boundary condition that v₀ = 0 means that v'₀ ~= 4.42 m/s.

Then v₁₂ = ( 4096 * v ) - v = 4095*v ~= 18100 m/s

This matches the numerical result.

 

[mfb]

A short version of the formula: $v_i=(2^{i+1}-1)v$ where $v=\sqrt{2gh}$
Why do we have $\ell$?

If the initial heights of the balls are h, 4h, 16h, 64h, ... above the top of the previous ball, you get a periodic system where each ball exactly hits the ball below it when this is at rest. This requires mathematical precision, of course...

 

[micromass]

Recall that I'm asking three questions, not only the speed.

1) Find a formula for the height in terms of n
2) What value of n causes the top ball to go higher than 1 kilometer
3) When do we reach escape velocity

 

[mfb]

Height is proportional to velocity squared, so $h_i=\frac{v_i^2}{2g}$ with $v_i=(2^{i+1}-1)\sqrt{2gh}$ => $$h_i=h(2^{i+1}-1)^2$$
This neglects that g depends on the height.

For (2) and (3), see jbriggs444.

 

[micromass]

Height is proportional to velocity squared, so $h_i=\frac{v_i^2}{2g}$ with $v_i=(2^{i+1}-1)\sqrt{2gh}$ => $$h_i=h(2^{i+1}-1)^2$$
This neglects that g depends on the height.

For (2) and (3), see jbriggs444.

Remember that only parts (2) and (3) say that $\ell$ is negligible, not (1). So there is a term $\ell$ missing. It's correct besides of this though.

 

[micromass]

Anyway, congratulations to jbriggs444 and mfb for their nice solutions!